Increment on “__toString”

Question

I am not sure what the title should be, but the code should explain it better:

class Group {
    private $number = 20;

    public function __toString() {
        return "$this->number";
    }
}

$number = new Group();
echo $number, PHP_EOL;
echo ++ $number, PHP_EOL;

echo PHP_EOL;

$number = "20";
echo $number, PHP_EOL;
echo ++ $number, PHP_EOL;

echo PHP_EOL;

$number = 20;
echo $number, PHP_EOL;
echo ++ $number, PHP_EOL;

Output:

20
20              <--- Expected 21

20
21

20
21

Any idea why I got 20 instead of 21? Even then the code below works:

$i = null ;
echo ++$i ; // output 1

I know Group is an object that implements __toString , i expected ++ to work with the string from __toString or at least throw an error

Answer 1

Why don't you just:

class Group {
    private $number = 0;
    public function __construct($number = 0){
        $this->number = intval($number);
    }
    public function __toString() {
        return number_format(++$this->number); // pre-increment
    }
}
$g = new Group();
echo $g; // 1
echo $g; // 2

I use something like this to format offsets in tables.

Answer 2

The order in which the operations happen is important:

1. The variable will be fetched as an object, it won't be casted to an integer (or something else).

2. This ++ operator increments the lval (the long value) of the zval, but does normally nothing else. The object pointer remains the same. The internal (fast_)increment_function will be called with the zval which has a pointer to the object, which checks for the type first. If it's an object, it does nothing. So when your zval is an object, it is as useful as a no-operation. This won't output any warning.

3. Only then the echo instruction performs a string cast on his arguments: The __toString method is called and returns 20.

4. 20 will be output.

Answer 3

This is actually more feasible than you may think -- just need to do a little bit more type casting, as follows:

<?php

class Group {
private $number = 20;

    public function __toString() {
        return (string) $this->number; // replace "" w/string cast
    }
}

$number = (int)(string) new Group();
echo $number, PHP_EOL;
echo ++$number, PHP_EOL;

---

You don't have to use the string cast of course in the magic __toString() but I personally prefer to read the code that way rather than seeing the quotes -- but I think that's just a stylistic preference.

Casting the newly created object as a string causes the magic __toString method to automatically execute and it returns a numeric string which when cast to an int allows you to display the number, increment it and display it again.

Incidentally, the space between ++ and $number is okay; I closed it up b/c that is what I'm used to in other languages like C.

Answer 4

To answer you question with a little bit of code.

$number = new Group();
echo gettype($number);

$number = "20";
echo gettype($number);

$number = 20;
echo gettype($number);

Will result in

object
string
integer

The three cases:

• You can't do any integer operation on a object, that why your code does not do what you expect. The __toString method will called very late, when the acutal output will computed, after you unsuccessfully tried to do an math operation with it.
• You can to math with strings, because PHP internally converts them back to numbers
• Obviously you can do math with integer

Bonus:

This will work:

$number = new Group();
echo 1 + "$number"; // 21

It converts you object into a string, which could be converted into a number for a math operation.

Answer 5

I think it might be clearer with just changing the names of the variables like this :

class Group {
    private $number = 20;

    public function __toString() {
        return "$this->number";
    }
}

$group = new Group();
echo $group;//print 20 as per your __toString function

++ $group;

Now it seems obvious : what is supposed to do a '++' operator on a object of type group ??